Approximate straight-line mechanisms through four-bar linkages

Approximate straight-line mechanisms through four-bar

linkages

Citation for published version (APA):

Dijksman, E. A. (1972). Approximate straight-line mechanisms through four-bar linkages. Romanian Journal of Technical Science : Applied Mechanics, 17(2), 319-372.

Document status and date: Published: 01/01/1972 Document Version:

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APPROXIMATE STRAIGHT-L1NE MECHANISMS THROUGH FOUR-BAR LlNKAGES

lIy E. A. DIJKSMAN·

...

Appro'llimate straight-Iine mechanisn{s. malnly. are takins your aUention In thls paper. The meehanisms considered are four~har IInkage, or special forms of these, such as the slider-erank. the Invertf'd ,S}lder-erank and the Inversion of a Seotch-yoke mecbanlsm, wbere tbe yoké is the frame (t'lIIpsograpb).

All me('hanlsms looked upon have coupler-points attachcd to tbe coupier. the con/3ecting rod or to tbe slldcr of tlle invertetl slider-crank mechanlsm. Tbe-eoupler-points deseribe coupler-curves wUh approximatc stralght-parts. In the dl'slgn-position of the mechanlsms the tangent to the curve has a four-, f1ve- or even six-point conta('t ",ith the curve descrihed. In all cases obser'\red the geometrie design of the mechanisms is a simpil." one. although the kinematic derivatIon behlnd is not. Generally. more dcsign-degr('('s of freedom are taken if a higher-point contact of the ('oupIer curve w!th the tangent is wanted. But in most cases sufficient deslgn-deli!\'ees of fn'edom are left to tlle designer to meet other requirements not eonsldered 111 thls paper.

Special altention is given to partieular de~ign-posltlons of the IInkag!'s. In thos!.' cas!'s th!' "eometri(' design of th!' IlH'('hanisms is gt'lting even mor!' simp)!' and an oVt'fall pictul't, Is acquh'ed from whkh :l designer may piek his cholee.

1. KINEMATICS OF THE PLANE MOTION

Tbe relat.ion between the polar co-ordina·t,es of a point Xo('o, cp) of the path and its oorresponding center of curvature is given by the equl\-t.ion of Euler-Suyary

] _{1 1}

(1)

"

3 sin I'?

Here 1.be po]e P is the origin of co-ordinates, and 3 the dia.meter of the inflect.ion circ]e, which touches the pole-tangent p in P (see Fig. 1),

8 a]ways being positive. Xv being the intersection

(=F

P) of pole-radius PX and tbe inflection circle, cp is measured 80, that 4. XI/,Pp = CP ..

>

O.

• Eindhoven University of Technology, The- Netherlands

320 2

The ~relation bet ween theradii of curvature of tbe centrode (Ro) and those ofthe 'moving centrode R is given by t,he equation

1 1 1

- - - = - .

B . • BoS (2)

In Fig. 2 tbe situation at tbe observedmoment is indicatedwitbout, and thc one af ter dt 'SOO. witb anaccent.Àt the moment of obser,vation tbe

n

Fig. 1 Fig. 2

moving 'plan rotates on tbe pole rPwitb an angularvelocity of <u' rad/sec,

80

XX'

= r<udt. dt sec, later, tbe pole coincides witb P'. Suppose

PP-'

=

= ds. Projecting P'X' on PX now leads to tbe equation

dr = - ds cos lp. (3)

Projecting P'

X'

on

1.xX'

gives thc equation

(r + dr) sin (dIP +dq,) = r<udt +ds sin lP. (4) Bearing in mind, that the diameter of the inflection circle by definition satisfies the equation

.S

=._

dB/dt

<U

andthat Bo dq, = ds, (') with (2) can be written

dep =.,sin ep

_.!.

dB f' B

(5)

(6)

3 STRAIGHT-LINE MECHANlSMS :l21

(1) leads t.o the equation

rl

p

=

ro - r

= .

.

8 sm cp-r (8)

80 with (S) and (7) we have

dP=S82p2(o)sincpcoscp(_.!..+

1

+

1)

(9)

fit r2 _{r l sin cp m cos cp}

where S 1 1

-=-+-

_{l R 8} S8 m= -d8/ds (10) (11)

The locus of those points of the path, where the radius p of curvature haR a stationary value, is caUed circling-point curve and is indicated by k ... It is found by the condition dp

=

o.

dt So we have

.!=

1

+

1

l sin cp m cos cp r

which is in cartesian co-ordinates

(;C2

+

y2) (mx

+

1y) -lmxy = 0

This (',an be written in pa,rametric form

Kz = x·

+

y. - ~x

=

0 }

Kg

=

x·

+

y. - Ày

=

0

where ~ and À answer the equation

~+~-1

O. 1 m (kit) (12) (ku ) (IS) (14) (14 a)

322 Je. À. DJJKSMAN

~---~---.---~---~

The point X* with co-ordinates (1', À) then lies on a straight line, inter· secting the abscissa. Mis in Mand the ordinate in Land wh ere by PM

mand PL

=

Z. I' and À are the respective diameters of circles K", and Kw, which intersect outside P in a point of ku (see Fig. 3). From this a design of points on k .. can be deduced (see Fig. 4). The foot X. of the perpendicular, dropped from the pole on a diagonal of the rectangle, 2 sides of which respectively coincide with the abscissa axis and the ordinale of point X*(IL, À) lies on k ...

Fig. 3 Fig ...

A possible form of kil. is given in Fig. 5. The locus of the oontres of curvature, corresponding with the points of the path k .. , is called

centering-point curve and is indicated by kIS' lts equation is found by eliminating llr from (1) and (12)

1 1 ro

=

-'o-s-in-IP

+

m cos lP 1 (15) where 1 1 1

---=:-.

I

'0

3 (16)

Just as we have a design for kw. with the straight line ML, there is a si-milar one for k .. with the aid of the straight line MLo (where PLo 10

and

.it

P

=

m). Centering-point curve and circling-point curve have a simila.r shape; büth curves are shown in Fig. 6.

STRAIGHT-LINE MECHANISl\IS 323 The point, where the path tangent has a four-point contact, is called Ball's point. This point lies on the inflection circle as weIl as on the circling-point curve. Generally spea.king there is only one such a point, since it does not coincide with the pole.

focal axis kil /I asymptote kil

'\.

\

\

\I

Fig. 5

lt is also called ufI,dulatwnpQint and is indicated by the letter U.

U lies on k.. Rence the corresponding centre of curvature U 0 lies on kIS'

BaU's point a1so is a point of the inflection circle, 80 U_o is a point of kIS,

lying infinitely far away. As U, U_o and P are lying on one straight line, this means, that Ball's point is to be found at the intersection of the line througb the pole in the asymptotic direction of kIS and tbe inflecf.ion circle.

324 E. A. DUKSMAN ₆

We find the asymptotic direction by equaling zero (Xl

+

,1) (mx

+

ló1l),

which is the term of the highest degree of the equation for kG in Cartesian co-ordinates. Hence the direction-coefficient of the asymptote is - mllo.

"1.0

Fig. 6

2. DETERHINATrON OF lAL L'S POINT IN ANY POSITION OF A FOUR-BAR LlNKAGE In a given p08ition of any four-bar linkage, we can design tbe pole tangent and the pole normal, with the aid of the theorem of Bobillier (see Fig. 7). As A and B are points of the circling-point curve k .. , the cor-responding centers of curvature Ao and Bo lie on ka. Point Ao now is asau-med to be the foot of the perpendicular PAo, dropped on the hypothenuse of a right-angled triangle, of which the sides containing the right angle coincide wUb 'I, respectively with n. The angular point A: opposit.e to P

7 STB.AIGHT~LINE MECHANlSMS 325

of a rectangle, of which two sides coincide with the above-mentioned sides of the triangle, now can be determined. The point

B;

is fixed in a similar way. As may be already known, the straight line

A; B:

then intersects the pole normal in Lo and the pole tangent in M. Hereby is P M

=

mand

PLo

=

Zo. \ \ n Lo , P :GUp/er curv~ " , , Fig. 7 I " ' M po

The perpendicular PQo in llPMLo so has the direction-coefficient

+nt/l_o. Mirroring PQo in the pole normal gives a straight line with thc direc-tion coefficient - _nt/lo, which intersects the inflection circle in Ball's point.

The points Aw, Bw and P lie on the inflection circle, whereby PA 2 ₌

=

AAo'

XA

tD and pJj2 BBo' BB,,, so the inflection circle can be de- .

signed and with it Ball's point.

Choosing the coupier point K in Ball's point and realising coupier triangle A BK, the coupier curve traced by K, has at least one Ba11's point and the motion of the four-bar linkage gi"'es us a proper straigbt-line guidance.

.326 E.A.DIJKSMAN 8

3. SIMPlE RELATIONS &ETWEEEN THE ANGLES OF A COMPLETE QUADRlLATERAL AND THE QUANTITIES m, I, '0' R, Ro AND 8 .

Be8ides by tbe lengtbs of the sides and the position angle of one

()f tbem, the quadrilateral also is completely fixed by giving tbe line part PQ and tbe 4 angles,tbe sides form with tbis line part. Tbe senRe ()f rotation is assumed to be positive when counter-clockwise, and tbat for tbe angles ~8 and ~2 from tbe sides to tbc collineation uis and lor llle

Fig. 8

".

Fig. 9

angles ~1 and ~3 from the collineation axis to tbe sides (sec Fig. 8). Tbe angles , .... and

'B

are also sbown in the same figure. In Fig. 9 is shown tbe determination of line ML for the circling-points .A and B.

STRAIGHT-LINE MECHANISMS ₃₂₇

On the ground of the theorem of Bobillier will be 4-QP A = 4-BPpt so

CPB

+

~l

7t)

and ('tan CP ... = - tan (3a = - ";'a,

and

CPA

+

~3 =

7t,

tan CPB = - tan ~1

= -

Tl'

With the a,id of Fig. 9 we see, that

while . a180

f l

PA

PB

PB 11:₂

=,

' Y 2 = -COS CPB sin CPB m =

_:e. -

x,.

!lt. - Yl With the help of Fig. 8 we deduce

(17)

(18)

(19)

PA COR CPI! PA cos ~1 EP EAltan "1 TaQA "'. QAIQP

_.

=

= =

=

= - =====

== - •

==

PB cos cP"" PB cos (3a DP DBltan ~a Tl QB ";'1 QBIQP

Ta sin ~l sin (~2

+

~a) Ta

+

Ta

- + - ~,

Tl sin ("1

+

(32) sin ~a Ta

+

1'1 80 with (18) and (19)

(20)

Changing A and B respectively by Au and Bo, we find in a similar way

(2] )

With (16), (20) and (21) we find

(22)

m

328

and from(20) and (22)

l = 1 _ TI

3 To

while (21) and (22) lead to the equation

1i)

(23)

(24)

With equfltNons(22) &nd (23)it is posswle to determine the imttmtaneom in'Oar>Îant'll of tàe terms on the left-hand. side jn a directly measurable way

fro·m a given set of points A, B, Bo and Ao. lVi th (10) and (23) finally we find the equation

and also R_To - T t

--

.

8 2'to

+

TI Re = TO - Ti 3 2Ti

+

TO

4. DETE1U'1INATION Of THE BUa.MESTER. POINTS IN THE FOUR.-BAR. LlNKAGE

The Burmester points answer the equations

(25)

(26)

From this it follows, that apath, traced by sueh a point, has a five-point vontact with the corresponding osculating circle of the obsen'ed moment. The last relation gives, with the help of equations (3) and (7) and

af ter substituting the vaIue for r from (12), an equation of t.he fourth degree in tan tp

mI d

(R)

mI

- - 8 - - tan tp

+

3R' ds 3 llo

o.

11 STRAIGHT-LINE MECHANISMS 329

llenel), tJw four Burme~ter points lie on pole radii with a direction coeffi-(·jcnt" I'u,t,i ... fying t his cqu3,tion.

Jf A and B Ill'e two Burmester points t,hen tan CPA

= -

"a and tan cpJl - ':'1 Ratisfy (27), so (27), with the shortening T=tan cp takes

tlw fortn

(28 a) .or

(28 b)

With (20), (21) and (25) equation (27) gives the relations

(29)

and

(30)

In addition t.o tbe centers of rotation .A and B two other Burmester point R may be found with the aid of the quadratic equation

: 11 ( 1 1 1 1 ) 1'1 1'a

tan cp

+

"I "s -

+ - - - - -

tan cP

+ -

=

o.

: ':'0 1'11 'I's 1'1 1'01'2

(31)

The two pole radii determined by this equation intersect the circling-point curve just at the Burmester circling-points we are looking for.

By comparing (28b) and (27) and substituting the values for 0 nnd

D, we find also

3 -

d

(m)

- = 2 - 1'11'S

[

1

+

(1'1

+

1'a) -

(

1 1 1 1) 1]

+ - - - + -

.

d8 3 1'0 1'2 1'a ':'1 1'01'11

(32)

330 E. A. DUKSMAN 12

S. ANGUlAR CHANGES (0,) AS A fUNCTION Of THE ANGLE (Bi) lTSELf

The left-hand term of (33) can be written

whereby

and

d'ro = (1

+

-rä)

l _ _ w

d!P20 1 - (d~2/d~o) (36)

Elimination of (d~./d~o) from (35) and (36) leads to

(37)

(37) in combination wUh (33) and (34) gives the equations

(3~)

and

:

).

'I

(39)

Tbc last equation can be also dedueed from the preceding by kinemntic inversion.

13 STRAIGHT-LlNE MECHAN1SMS 33t

As

1 _ sin ~o/sin (~o

+

~l)

sin ~,/sin (~t

+

~l)

= - 1 _ sin ~l/sin (~l

+

~~) = sin ~3/sin (~3

+

~o)

1 _ To Tl

+

Ta

Ta ';'1

+

TO Tl ';'a TI - TO

1 _ Tl Ta

+

TO

= -

TO TI Ta - ';'1

Ta Tl

+

TO

(40}

(see Fig. 10) equation (39), alter eyelie ehanging of tbe numbers of 4: links,. takes t,he form

(41}

20

31 10

Fig. 10 while (38) passes into

(42)

With the signs agreed upon for the angles ~m, the angle ~2' by cytlic ('han-ging, passeR into TC - ~3; ~o into TC - ~l; ~l into TC - ~2 alld ~3 into TC - ~o, 80 ":'2 passes int.o - Ta;

"0

into - Tl; ":'1 into - 7% llnd ':a into. - Te, etc.

332 E. A. DJ:.JB.SlIIA.N ₁₄

6. IAU'S POINT AS A BURMESTER. POINT

If the coupier curve has a five-points contact with iti' tangent, this point of contact not only is a Ball'g point but also a Burmester point. This will ba the case if - m

(=

T~

Ta)

is a root of the quadratic equation

lo '0

(31).

HIR

80 this equation gets the form

(tan lP - -;;; Tl Ta) ( tan lP - TI 1 )

=

O. (43)

(1,ompa.riDg (43) with (31) leads to

or

- Tl - Ta

+ - +

1

':'1 Ta -

(1 2)

+ -

=

o.

I

TI Ta TO

(44)

U the angles of the quadrilatera.l satisfy this oondition, Ball's point is

a. Burmester point. We speak of Ball's point with excess 1 [2].

Propo8ition. IJ BalZ'8 point coinciàeB with a Burme8ter point the coupler inter8ectB the pole normal in ti point T, BatiBJying the (:onditio,~

1 1 1

1==+=

-I PL P~ 2·PT

!---~

The proof of this proposition is given by Fig. 11. By the sine rule in 4 PTA will be

15 STRAIGHT-LINE lIIlECHANlSMS 333

As .A is a circling-point, it is true tbat

_1_= 1

+

1 •

P.A , sin (180° - ~a) m cos (180° - ~a)

Fig. 11 80

PT

= _

sin (~1

+

~II) • sin ~a cos ~a

Z cos (~

+

~2

+

~a) C08 ~a - (l/m) sin ~3

Substitution of the value - ~ for

'fm

gives af ter some rewriting, tbe equation

On tbe ground of t1l,e (B~) cofKlition tbis can be written

1 1 1

- +-

= - = .

l Zo 2PT (45 a)

wbicb bad to be proved.

16

7. BALL'S POINT WITH EXCESS 2

The coupIer curve has a 6-point contact with its tangent in Ball's point" ü the lelt-hand term of (44) possesses an extremti-m for' the Talue zero. Tbis means, that a1so the derivative to time of the 1eft-band term is zero.

80 a,part. from (44) we get the condition

o.

On the ground of equations (38), (39), (41) and (42) this gives, a.fter divid-ing by ":e 'rt ,the relation

Te - T t ! (1 1 1 ) ( Ta 2Ta) (1

+

Ti) - - -

+ -

1 - - - -

+

Ta Ta T. Ti 're !

(1 1 1 ) (

Tl 2":1 )

+

(1

+

Ta) - -

+ -

1 - - - -

+

TO Tl TI TI 're (46) So (44) and. (46) are the conditions for a Ball's point having exooss 2.

More than a 6-point contact is impossible, the coupier curve having the degree 6 and a curve of this degree intersects a straight line at most in six rooI points.

The two re1ations, (44) and (46), between the angles (l., (lu (ll and '(l., being necessary &nd sufficient fer the point of BalI ha ving excess

.2, just Jean us free choice tor one of the angles ~",. As only real vaIues for a.ngles (l ... , satisfying equations (44) and (46) are usabIe the cboice of a second ang1e is limited. In the following the ümitation laid on this ang1e, for which we take (la, will be further defined. The angle, which can he

chosen free1y, be ~l'

Do we agree, that

17 STRAIGHT-LINE MECJlANISMS ₃₃₅

---then (44) gets the form

(47)

while, a180 by this, (46) may be written as

(1

+

ti) (t, - 's

+

tol (t, t_{s -} 1)

+

(1

+

ti) (t_{o -} tI

+

t,) (it t. - 1)

+

+

(1

+

ti) (t_s

'0

+

t.) • (1

+

it ta)

+

2 (1

+

t5) (tl

+

ta - t.) = 0

... (BI.). (48)

Elimination of to from (Bl.) and (BI.) then gives

(1

+

ti)

[it - t_{s -} (t.t_s -1)

t.l

(t_at , - I )

+

+

(1

+

ti) [ta - tI - (it ta - 1) t.l (t. t. - 1)

+

In this equation the coefficient of

tI

is identical to zero, whieh leaves us a quadratic equation in t2 • Af ter working out the suceessive eoefficients

of this equation and dividing by the common factor 5, we get

(49)

As only real roots for ta count, it and t3 have to he ehosen in sueh a way, that the discriminant of the quadratic equation will be

>-

O. This will be the case, if

or if

336 18

This mea.ns

- (ti

+

1)

(ti

+

1) ~ _2(t1

+ (

_{8) (tl} t₃

+

1) ~

(ti

+

1) (ti

+

1) (50) The nght-hand inequality can he reWTitten to

[

t _

ti+l

]2~ 4t1(tr-t1+1).

3 (tl -I)',;?' (tl - 1)4 .

If ti

<

0 this will always he satisfied, because in this case the right.-hand

term will he negative.

H, however

'1

>

0, tben should

I

t 3 - (~

t~

_

+

I)' 1

I"'"

- ? (~ _ 2 1)'

Vt

1 (t2 1 - t 1

+

1)

The

left-1umiI

ineqnality may he rewritten to

If tI

>

0, this will :dways he sa.tigfied. If, however ~

<

0, then sbould

I

t 3

+

(~

ti

+

+

1)21 1

I

~

,;?' (~

+

2 1 ) , 1

V~

t (t'J. 1

+

I "1

+

1) .

Summed up, in case ~

<

0 must he

t_{3 :;).} 1 [2

V -

~

(ti

+

~

+

1) -(tr

+

1)] - f( - ti) (501$) (tl

+

1)1 or - 1

V

.,

t8~ [2 - t1(ti+t1+1)+(tï+ 1)]=-g(-It) (tl

+

1)' (50b) and in case

"1

>

0 t3 :;). 1 [tl

+

1

+

2

V~

(tr - ti

+

1)] = 9(t1) = 9

(~)

(tl - 1)2 ti (50 c) or

19 STRAIGHT-LINE MECHANISMS 337

In Fig. 12 the two forbidden areas have been indicated, the values for

'1

and 13 being measured along the axes.

To design the quadrilateral we proceed as follows : Choose tI = Til

and la -:81 _{in a,ccordance with the permitted yalues of Fig. 12. Then we}

Fig. 12

find 12 -:it, with the help of the quadratic equation (Bl2 ) in th is quan-tity. For one of these values t2 we determine the quantity '0 -:ö1 with the

(Bl1)-condition. Now we can draw the quadrilateral wit,h the aid of the 4 values tI'

'2' '3

and t_o, being U = B12 (it may be noticed, that any point

"".1

~." "4!.t"';

, .... ir

Fig. 13

of the area of Fig. 12 corresponds with 2 solutions of the problem). In this way can be realised, it is true approximate, but still very good

338 E. A. DIJKSII4AN ₂₀

a. STR.AIGHT-LINE GUlDANCES WITH fOUR-BAR LlNKAGES IN A SPECIAL POSITION

Tbe design of a four-bar linkage, with coupier points tracing a.n approximate1y st,raight path, will be much simpier, if tbe linkage is in a special position.

Besides, by submitting successively sueh positions to a olose exa· mina.tion, we get a good view of the possibilities, presented by four-bar linkages . .All tbe above-mentioned positions can be obtained in positions of the coupier plane, where either tbe circling-point curve or tbe oentering-point curve is degenerated. Tbese are positions, wbere two links overlap each otber partly, lie in each others produced part or are parallel and also in positions, wbere tbe collineation axis PQ is perpendicular to one of the four links.

8.1. THE CYCLOIDAL POSITION OF THE FOUR-BAR LlNKAGE

If tbe diameter of the inflection circle reaebes an extremum, tben the coupier plane is in a so-ealled cyeloidal position. Tbis bappens wben (d8/d8) = 0, 130 also m -1

=

o.

Tbis results in k" falling to two pieces

1. cos lP

=

0 in other words ft

=

1t/2, tbe equation for the pole-fWrmal,

2.,.

= l8in <p, the equation of a cirele c-k" througb P witb center point (0, l/2).

Fig. 14

Also k. fa1ls to two pieces, namely the pole-normal and a circle, c - k. through P with center point (0, 10/2). (see Fig. 14). Tbe 3 circles

21 STRAJGHT-LINE 'MECHANJSMS 339 c - kv, c - 1.:0 nnd the inflection circle are interdependent through the relation

1 1 1

- - - = - .

1 lo 3 (16)

A design, based on this relation, is given in the same figure. Ball's point here coincides with the infJection pole. Apart from folding linkages, we dist.inguish 3 cases:

Case A: Both the tu,mi1/,g pointa A ancl B Zie on c - k,. (and conse-quently Ao and Bo on c - l.~a)' Figure 15 shows us the four-bar linkage in

IC,

340 22

the posit,ion with the coupier parallel to the fixed link (AB! !AoBo). The collineation point Q here lies infinitely far away, 80 TO = ÀT2 = 0, where on the ground of (25) À = (R!3)

+

1 =F1, because a180 R!3=F O. 80

1 - 2(R!3)

the position may be obtained by substituting _{ÀT 2} for .. _o and tIJen decrease

TI to zero, keeping R! 3 constant.

This transition to the limit, genel'al1y speaking, results in

rn ( TO

't"t)

- = - ":'1 '3 -+ 00 3 "0 TI - = - -+00 l TI 1t~ = _ Tl Ta

'0

TO

rema,ining finite alld

'0 ":'0 - ":'11

- = 3 ':'1

remaining finit.e.

80 the quadrilateral inde<ld is in a cycloidal position. The design of the level-luffing crane, as shown in Fig. 15, is an example of tbis. Other known examples are the straight-line guidance of Tschebycheff and that of Roberts, as shown in Figs. 16 and 17.

Do we demand a better nestling of the coupier curve against itl'l tangent than Ball's point offers, the

B'lt

condition (44) must be satisfied. In t,his case there is ar five-point contact with the tangent .

. After substituting ÀTI for

't"o

in (44) and tben decrea8ing ". to zero,

remains

O. (51)

By the expression for À and equations (17), this can be written

1 (.

R) ( R)-l

"

0:-, \ ....

"

.... ,

"

" "

_"

"

p

/

---" _{" , ,} ", , , 16 p Plg. ti 341

342 E. A. DUKSlIAN ₂₄

GeometricaJly this equation can be realized as folloW"s (see Fig. 18): a. Draw p, ft, C - kv and the inflection circle,

- 2

-b. Determine c - ka with the equation PL

=

_LW.LLo,

c. Choose B on c - ku '

d. Intersect PB wUh c - ka in Bo =1= P,

e. Draw UBI/BB_o,

f. Determine point S

(0, : 8).

Fig. 18

g. Draw SB/lp and determine intersection R of SR wUh URt

h. Join Band L (0, Z) and intersect BL with c - k" in A =1= L,

i. Draw pole radius PA and intersect PA with c - ka in Ao,

STRAIGHT-LINE MECHANISMS ₃₄₃

P.fuof

.!

cot 'tiE

=

SR

=

(~

a -

l)

tan (180° - IPd)

=

(I -

~

3) tan 9.""

2 2 . 2

SO Oll the ground of the ,siven design

l I

cot 9.{ . cot <PB = 2

a -

3!.

(52 a)

a condition, which can be reduced to equation (52) by (10).

To check the described design, it has to be shown, that, also by (45),

. . ( (1/2) 1 )

coupIer .AB mtersects the pole nonnal n. at pomt T 0, 1 . 2 - ( 13)

For a 6-point contact with the tangent conditions (47) and (49)

have to be satisfied. The first led to (52). The sooond, with 1'2-1

=

":2

=

0,

gets the form

(53)

We distinguish two casef\

Aa· ~ ta

+

1 0 (54)

A.s besides, on the ground of (51)

~

ta _1_ = - (1

+

~),

is

Tl Ta À

here 0

=

),-1 1-2(R

l

a), and so (RIa)

=.!.

On the ground of (10)

1

+

(RIa) 2

this will be the case if l

=

a.

The circle c - k.. coincides with the inflection circle, while c - ka falls apart to the pole tangent and the infinite far straight line. As points A and B are to be found on c - kOl , the coupIer executes an elliptic fMtwn.. The coupIer point, chosen in Bl.,

which ooincides with the inîlection pole, then even traoos an exact straight !ine. In Fig. 19 this is demonstrated in case PA J. PB, which is the geometrical meaning of (54).

(55)

This will he the case

344 E. A. 'Dl'.'JICSMAN ₂₆

There will be a. ~-point contact with the tangent in the point of BalI, if

as 'weIl cot !P .•. cat !PB = 2 ; - 3

as cot !P..4+cot !PB = 0

"

Fig. 19

Geometricall)- this can be realized as folIows (see Fig. 20) :

a. Draw t,he pole t::mgent, the pole nOl'mal ànd the inflt'ctiou ('h·(·ll'~

b. (,hoose the pole radius P.A and see, that the pole radii PB ancl PA

ha"c mirrored themselved in the pole normal 11"

c. Draw UR/lPB,

d. Fix. point S

(0,

~

a).

e. Draw SRllp, interseeting U R in R,

f. Drop thc perpendicular from R on P.A and let .A be

nte

foot of this perpendicular,

g. Draw the circle c - kil through Pand .A, with its center on '11"

h. lnt,ersect c - ku and PB in B,

i. Design c - k .. with the help of (16),

j. Intersect c - ka with P.A in .A_{o nnd with} PB in Bo, k. Draw à.ABK, whereby K = W = U = BI".

Thc checkpoint is again .AB intersect,ing the pole normal at point

T

(0,

'1

2 ). It is possible t.o choose the ratio

II

a

in su eh a way, 2 - (lIS)

27 STRAIGHT-LINE MECHANISMS 345

Fig. 20

The ca,se being it bas to be T

=

K = W Bil, and so 1/2 - 3.

2 - (lla)

From tbis we find

~

=

i ,

llP AB is equilateral.

3_4_6 _____________________ & __ A_._DU8S~~MAN==~ ___________________ ~2S

The corresponding meehani8m has been drawn in Fig. 21.

11

---~~~~~~---,

Fig. 21

The remaining degree of freedom ma.y a180 be used to 10eate the coupier point on the fixed link AoBo (see Fig. 22). In this case will he

PT PB _{- , and 80} l

29 STRAlGHT-LlNE MECHANlSMS 34ï

=---.---'Vith (16) thh! leads to aquadratic equation in 1/3

lt l

2 -7 --

+

4 =

o.

at

3

'We find 2 solutions

( ; l

2,781 (Here the line through T parallel to p intersects the circ1e c - k .. a.t 2 complex points.A. and B, which means there is no real rnechanism corresponding to this solution).

and (

~}~

=-

0,719 (see Fig. 22).

Fig. 22

Case B: Turning point A lies on n and B on (l - kv' On the ground of

t.bc theorem of Bobillier, the collineatiott axis PQ, in this position of the quadrilateral, will be perpclIdicular to B~. (sec Fig. 23). This means, that Til

=

ta

=

O.

:348 E. A. DIJXSMAN

Just like in C&se A,

m - = -+ 00, -+ 00, l 3

TO - TI remains finite and that TO ZO TO-T _ = IJ remains finite. II TI Fig. 23 30

:u STRAIOHT-LINE MECHANlSMS 349

80 the quadrilateral is indeed in the cycloidal position. The so-called unsymmetrical straight-line guidance of Roherts gives us an example of a quadrilateral in this position (see Fig. 23). Here, the straight guided coupIer point is in a Ball's point.

If this is a Bll - point, then will he by (16) and (45), with T = A

--

_PA=

_----"1_2_

2-(1/3) ( Bl:t). (57)

In figure 18 has been indicated the way of fixing intersection T. The same design may be used to determine the point A = T. This results in the following design (see Fig. 24)

a. Draw p, u, c - k .. and the inflection circle,

b. Determine c-ka with the equation PLt. = LW. LLo,

c. Ohoose B on c - k.. and interseet PB with c - ka in Bo =1= P,

d. Draw URIIBB01 e. Fix point

s(o,: 3),

f. Draw SRllp and fix intersection R of SR and U R,

g. Join R to L (0, I) and interseet RL with c - k .. in A' =1= L, h. Interseet A' Band u in point A,

i. Fix Ao on n with the help of the equation PAt AAo . AA ... ,

wbere At!' W, being the inflection pole,

j. Draw b.ABK, whereby K = W = U = Bl:t.

A necessary condition to he fullfilled for a 6-point contact with the tangent is the (Bt~)-condition (49). Fm; ta

=

°

this leads to the equation

(58)

Again we distinguish 2 cases

From (47) it follows with ta = 0, that 2t_o

=

t1 - t2, and so also to

=

't'ö1 _{0. By (23) this gives I}

₌

_{3, and so c - kv coincides with the}

inflection circle and c -klJ falls apart to the infinite far straight line and and the pole tangent.

From (56) it follows with 1= 3, that A will he at the center of the inflection circle. The mechanism, coming up to all tbese demands, appears to be an isosceles slider-crank mechanism (800 Fig. 25).

350 32

11

Fig. 24

STRAlGHT-LINE MECHA.NISMS 351

Coupier AB then executes a so-called elliptic motion, coupier point

K tracing a ,traig'" line through Ao.

Bb. t1t.-1

=

O.

Pram this it follows, that ';1';. = 1, the geometrical meaning of

which is QA..L PA (and so AB..L AAo).

À design, for which this is true and where U = Bl., is the following (see Fig. 26):

a·. Dra:w p, n, c - k.. and the inflection circle, b. Design c - kil on the ground of equation (16), c. Fix point A on n with co-ordina.tes

(0,

1/2 ) t

2 - (1/3)

d. Fix Ao on n with the aid of the equation PAl = AAo • AA." where A., = W, being the inflection pole,

e. Draw a line parallel to p through A, intersecting it with

c -

kil at point B,

f. Interseet PB with c - ka at point Bo

=1=

P, g. Draw A ABK, where K = W = U = Bli •

The design further spea,ks for itself.

It is possible to use the remaining degree of freedom to locate the point (BI,,) on a line through Bo and perpendicular to tlte crank AAo

(800 Fig. 27) We find again l = - - - = - ,

PK

PBo 10 so 1 -~-=-. 8, 2 - (lj8) Zo . and so with (16)

( ;)] = 2,718 (lending to an unreal

m~hanism)

and (

~

t

= 0,719 (800 Fig. 27).

In the la~t <,ase point A lies on the pole normal and it has co-ordi-nates

(0, ',:),

Point B lies on c - k. in sueh a wa·y, that AB

..L

AAo.

Case C. !ffJ,rllillg poillt BlieB 011. n 411d A Q1I, c - k •.

Bere is Tl ~ 00.

The only difference between this case and the former one is that the links 1 and 3 have been ehanged mutually.

-'52 , ,

,.

, ! I I I ,

.-V'

: ' \ Fig. 26 .' • Fig. 27

35 STRAIGHT-LINE MECHANlSMS ₃₅₃

B.2. THE roUR-BAR LlNKAGE IN TI1E POSITION. WHERE R 2Re •

On the ground of equations (2) and (10) this position will be l'e:1rlted in case l-1 = 0: The circle point curve here also falls apart to two piere~:

1. sin cp

=

0, 80 cp

=

0, the equation of the pole tangent,

2. r

=

tn C08 cp, the equation of a circle C - k .. through P with thc

~nter point (tnJ2, 0).

In this ca,se the centering-point curve does not fall apart. Ball's point is the intersection of c-k .. with the inflection circle, not coinciding with P. Excluding the folding positions, there are 3 different cases to he

disUnguh;hed

Case A. T1H~ turning point A lies on c - k .. and B on p.

On the ground of the theorem of Hartmann, the center of curvature of the turning point B coincides with the pole, so in this position the ('rank AAo lies in a straight line with the fixed link (see Fig. 28).

130 in this case will be TO

I

1

10'

IJ. Tl

=

0, whereby IJ.

= - -

":'3 •

nt

Using this with the equation of the instantaneous invariable..;, Wt.' ha,-c in gi'uel'tll, th at

..i

= 1 -

..:::!. _

00 and that!!1'-= _'I."IT3 remains

a

T O lo '0

fillit-e. So tht> qua.dril~Ltel'i11 is in the position where R = 2Ro. Au eX<lmple of a mechanism, for which this position has been chosen as a starting point, is the straight !Jlû(lallCe of Hoeeken (see Fig. 29). As a particularity the turn-ing point B ba." been chosen in the center of c-ku and the turnturn-ing point B on the joining line from B to U. In consequence of this the coupier point, being Ba,}l's point lies on thc produced coupier. The con'esponding coupier eurve is 8~-mmet)'ic.

With ':"Q

=

W:'ll the condition for a five point contact becomes

or with "I - 0

1

1;0 wit,h t.he term for fL

1 m - == ~3

+2-':"2 lo '1."1'1."3 2'1."3 = 0

"2

fL (;j9) (60)

• Thou~h t JU' ratio of tIl(' dialllcll'l' of thc rollcr circlcs is thc same as that of th~ Gardioid m(·tioll (:1"0 ('alkd inwrh'd ('anlan mation). wc only speak ot a cllrdioid positioll ot til('

tourbnr linlwg,' if. in :Hldil ion to tilt' ,::Î\'('I\ ratio, thl' diameh'r of thl' infll'ctioll drdc has l ... rlll'd

I

I

r

\

355

By.(24) tbis alsomay be written

(Bit)· (6U

As may he known AB inter8ects the pole nonnal in ft point T, for which (45)

As To = 0 will be on the ground of (24) Z. = - 3 and by (28) l-1 = 0,

- 1

so PT = - - 3. 2

Geometrically this means: that AB inter8ectB tAe pok normaZ in

tAe cenÜlr of tAe return circZe (see Fig. 80). 80 in this case :Qall's point is

ft Bl,. point.

A necessary condition for a siJ: point contact with the tangent

is a (Bl,)...condition. (49)

This may al80 be writ.ten as

"ïth Tl

=

0 this leads to

t .. ~ - (1

+

4)

la

+

t, = 0 or

(68)

There are 2 carieS to be distinguished

Aa.

tJa

-1 = 0 or ,;,,';', = 1 (64)

Geometrically this means PB..L Q B. 80, as point P coincides with Bo and Q with A, we have BBó..L AB or p

..L

AB.

Then the coupier intersects the pole normal in the improper point of n. This point does not coincide with the center of the return circle. 80

this case is an impossibility.

356 Eo A. DIJKSIIIAN ₃₈

This will be the case if the turning point B lies at the c.enter of c-k".

Sa there will be a Ball's point with excess 2, Ü AB goes throu,gh th.e cImter of th,e return circle and B coincides with the center of c - Kt< (see Fig. 31).

,

1 \ 1 \

,

\ J \ I \ I \ 1 \ 1 \ 1 ' I \ I \ 1 \ 1 \ 1 \ I \ I '. 1 \ I \ 1 \ 1 \ I \ \ \ \ \ Fig. '30

Case B. The turnillg point B lies on c - k" and A on p. Now ":'0

=

11' Ta O.

This will be the case ü BB. is in a direct line with the f~ed link.

By interchanging letters A and B, we get again the case discussed under

8.2.A.

Case C. Both the tuming point8 A and B Ue on c-k .. (see Fig. 32). As on the gl'ound of tbe theorem of Bobillier ~ APQ ~ MPB and

STRAlGHT-LINE M:ECHANlSMS 357

w

/symelricllllCis

f.C

Fig. 31

hesides

<r

PAQ

=

<r

P M B, we have aAP B,...., aAl P Band so

<r

PQA =

=

<r

P BM = 90°. 80 the position will he reached, if the collitteation axis-is perpenditmlar to the coupler. From this it follows, that Ta -;. 00. By

(23) then will be indeed 1-1 _{= O.}

From (44) it follows, that Ball's point has the excess 1, if

1 1 2

.31>8

-.so with (22) E. A. DlJKSMAN m ":1+":a=-3/2 As ":, ~ 00 we have by (24) Z. = - 3. Fig. 32

Besides I-I = 0, so (45) will get the form PT

lS tbe intersection of AB with n.

40

(61)

1

- - 3, whe:re T 2

80 here also the proàuceà coupler goes thrO'Ugl~ the center of tAe return cirele (see Fig. 33).

A necessary condition for a sa-point contact witb the tangent is

t.he (BIl) oondition. (49)

With '11 = ,,:;1 this gets the form

I

(~

+

tal

(~

ta

+

1) = 0

I

(68)

We distinguisb 2 cases

-41 STRAIGHT-LINE MECHANISMS

On the ground of (66) we now have TÖ1 = 0, go PQ...L A.B •.

Besides T:rl = 0 and so PQ...L AB.

Fig. 33

359

The qua.drilatera.l then is a .foldetl one, whioh will not be further

tDVestiga'ted.

(69)

Geometrioally this means PA...LPB.

80 there will be a Bl,-point if AB goea through the center of the ,etftrn

360 E. A. DIJKSMAN ₄₂

---Fig. 34

8.3. THE FOUR-BAR LlNKAGE IN THE POSITION WHERE Rij = 2 R·

On the ground of equa.tions (2), (10) and (16) this position will he

reached in case

lil

= O. The centering-point curve ka degenerates .in the pole t.angent pand a circle through P, having its center point on p. The asymprote of ka now ooincides with p ; a.nd so does the line through P in the asymptotic direction of ka. 80 Ball's point here is the intersection of p and the inflection mcle. This intersection is the pole and so U = P. But the pole always is a return point: IK) here it is a return point in the stra.ight

• Though the ratio of the diameter of tht' Tolkr circlt's is thE' sam!! as that or the cardan motion. also callt'd eJliptic motion. W~ do not speak of a earèan position of tht' four-bln Jinkagt' untU thc diametE!r of thc inflection circl(' has r('acht'd au extn'mum.

STRAIGHT-LINE MECHANISMS 361 , -guided path, which in g .. :meml is not desirabie (see Fig. 35).The position will be reachl'd, if the coupIer is in a direct line with one of the other two moving linkt; or if the collineation axis PQ is perpendicular to the fixed link AoBo. B~' kinematic inver8ion, we get again th(> case discusse<l undcr 8.2.

Fig. 35

8.4. THE CARDAN POSITION OF THE FOUR·BAR LlNKAGE

A combination of the possibilities, as discussed in thc foregoing, appears if, in the cycloidal position of the four-bar linkage, the ratio of the radius of the roller circles Ro/R = 2.

This will be the case if m

-1

=

Zo-I

= O. In this degeneration c - k"

coincides with the inflection circle and c - ka is infinitely large, so kil falls apart to p, n and the infinitely far straight line. So here any peint of the inflection circle is a point of Ball.

An example of a mechanism in this position, is the mechanism of EfJans, where the coupIer point has been chosen in the inflection pole (see Fig. 36). Any point of p may be the fixed turning point, as the moving turning point, coupled with it, coincides with P. Excluding thc folding linkages, the cardan position appears, if la. 'tl -'jo 00 and 't! = V Ta = 0 (800 Fig. 37). or if

362

Fi~. 36 ft

45 STRAIGHT-UNI: MECHANISIIS

363-lb. 't'a - 00 'and 't'1 = v' 't'l = 0 (see Fig. 36), the links 1 and 3 being'

interchanged. 80 it is not a limitation ü we restrict ourselves to case la.,

We find the geometrical meaning of the factor" by starting trom the cycloidal position,w ith PQ

i.

AoA and so 't'l -+-CX'I (see Fig. 38). By the

sine mIe in AP BQ, will be

For the observed cardan position will be Q Bo and 't', = V • 't'a

=

0,.

and 80 PB

=

+ "

('l0) PBo 1

+

v and al80 1 ' 1 =

+ - - - -

(71)-(PBoIPB) -1

The llleaning of the used symbola may be taken from Fig. 39. Unlike the positions discussed already, it is not the question here, which condition the quadrila,taal has to satisfy for Ball's point being a Bnrmester point but we have to aak which of the infinitely many undulation points on the inflection circle is a B~-point.

Formerly we had to find the situation of Ball's point in a fixed posi-, tion of the linkage, now we have to find the location of the Blt-point on the inflection circle. This point being a Burmester point, ean be found by-(31) with the help ofthe quadratic equation

As Tl - 00 and ". = V· 't'a = 0 it is true for the finite root, that

(.!. -

1)

tan IfIBI

+

_1_

=

0,

and so

'\I 1 V • 't'o

~

= _

11't'0 •

1 1 - v (72}

80 a pole radius, drawn in the direction given by tbix equution" intersects the infleetion circle at the B~·point.

36\

,

Fig. 38

.?_---47 STRAIGHT-LINE MECHANISMS 365

Substituting the values for v and TO leads to

(73) (PBo/PB) - 2

In this equation will be (PBo/PAo) >0 if ~o >0 and (PBo/PAo)

<

0 if ~o< O. Besides (PBo/PB)

>

0, if Bo and B lie on", on the same side of Pand (PBo/PB)

<

0 if Bo and B are separated by p,

In Fig. 39 has been designed, with thehelp of equation (73), a mechan-ism in the cardan position, the coupIer point being a point of BalI with excess 1.

U, particulaily, turning point B coincides with the center point of the inflection circle, then will be Bo

=

Wand v

=

1, and so B~

=

W*,

If, in addition, the center of pi\tOt on frame Ao is the improper point of p, t.hen the ratio YBI, will be indefinite and a",y point of the inflection

3)BI,

circle will be a B~-point.

The corresponding mechanism is an isosceles slider-crank mechanism, its connecting rod executing a carda",-motio", : any point of the inflection circle describes a straight line through Bo, which corresponds with the foregoing.

The Bla-condition (49) shows us, in combination with (47), that with Bo =1= W, there can be only one Bla-point, if to = 't'ö1 = O. This will

be the case if Ao is the improper point of p.

From (72) it follows, that then the Bla-point coincides with the pole, the latter being an exceptional point of the inflection circle.

Generally speaking a scalene slider-crank mechanism in the cardan position has no Bla-point.

B.S. THE TRANSLATORY POSITION OF THE FOUR-BAR LlNKAGE

In this position both the moving links, coupled to the fixed link; are parallel (AAoI/BBo). 80 in this case the pole will be infinitely far away (see Fig. 40). Generally speaking the pole tangent will be found by mirroring the collineation axis PQ in the bisector of ~ APB. As this bisector divides side AB in parts, being in the ratio of the standing sides of llP AB, the collineation axis has to be mirrored in a line through Pao andthe middle of AB, in case P = Pao. As on the ground of the

theorem of Hartmann, the center of curvature of any point of the path lying on the pole tangent, always concides with Pand in this case P

=

Pao,

p will be a branch of the inflection circle .

• A~ in this case the coupIer point lies on Lhe coupIer, the coupIer curve will be symmetrie, 80 the coupIer curve has two B/l"Points, Iying symmetrically with respect to thl' axis of symmetry AoRo. Th(' angle, between the tangents to these points, is fixed hy Lhl' choicl' of point Ao on p.

366 48

Tbe infinite1y far straight line generally intersect,g the circle point curve in t.he two isotropic points and in an asymptotic point. To these is added now a double point, namely the point p .... For a cubicaJ. curve this will only be possible, if the whole infinitely far straigh line forms a

"

~é::.::::"...J.,..:::::::::::====~o::---~ ~'r~

Fig. 40

p

·'.C.·

asymplote 0"

orthogon,r "yperbola

fhrougn A and 8 which

·S a b'1lnch Of 1("

branch of kv. The remaining branch then will be ot the seoond degree, here a hyperbola, for the pole tangent being an asymptote of this branch.

11 a straight line intersoots the hyperbola in two real points, the middle between these two points coincides with the middle between the two in-tersootions of this straight line with the two asymptotes of the hyperbola.

On the ground of the theorem of Bobillier, the collineation point Q will be a point of the asymptote of the hyperbola, not coinciding with p. As this sooond asymptote has the direction of the pole norma.l, it will be perpendicular to p. So the hyperbola will be orthogonal. Generally speak-ing we cannot find here a tinite point of Ball, bespeak-ing the intersection ot kv and the inflection circle, as the hyperbola only has an asymptotic contact with the pole tangent (see Fig. 40).

B.5.1. The trlAslatory position of the tour-bar link.,e. in whlch the tWO standin. links are perpendIcular to the coupier

11 in the translatory positioll of the linkage, AAo is perpendiculair

to AB, thell AB coincides with the asymptote of ka' going through Q. As in addition points A and B are circling-points, the hyperbola fa11s apart to its asymptotes p and AB, and BO an.y poim of p

wm

66 a point of BaU.

49 STRAIGHT-LINE MECHANISMS 36i

An excmple of a linkage in this position is to be seen in the design of a special type of level-Iuffing crane, built by N. V. Figee, Haarlem, Holland a·nd llsed by Hoogovens, Ymuiden, Holland, to unload iron-ore from sea-boats. In this case a bridge crane has to pass with its outrigger

.~_--IC.

Fig. 41

over t.he level-}nffing crane, so the height of the latter is limited (see

Fig. 41).

A. second example gives us the mechaniBn~ of Watt, the straight guided coupIer point being the intersection of the pole tangent and the conpier (see Fig. 42).

In thc observed position there will be a point of the pole tangent, having a five-point contact with tbe tangent to tbe pat.h in thi~ lloint ..

This point may be found by starting from a four-bar linkage in the car-dan position, as discussed in the previous paragraph, choosing the coupier point in tbe Bl_i point. Applying tbe tbeorem of Bobillier to this mechanism

or'hogona\ hvperbola

Fig. 42

leads totwo possible variants, one of tbem giving us a linkage in the translatory position, in wbich the coupIer point, a1so will be a Bil-point (800 Fig. 43).

8.6. THE CARDIOIDAl POSITION OF THE FOUR-BAR-liNKAGE

Tbe mentioned position is a cycloidal position, where the ratio of tbe radius of the roller circles RIRo = 2. 80 here will be .",-1 l-1 = O. In tbis case ku (!onsists of 1', '" and tbe infinitely far straigbt line, while klJ bas !allen apart to '" andtbe retmn circle. The point of BalI coincides with theinflection pole (see Fig. 44). Here will be ';'3~ ex> and '0

= .

=

IJ.

"1=

0, Ao being tbe center of tbe pivot, on frame, whieh has been chosen on the pole normal.

On tbe ground of the things discussed in section 8.2A, the point of BalI in tbis case will be a point of Burmester, if the turning point A coincides with the center of the return circle (see Fig. 45).

From S.S.Ab it folIows, that the point of Ball will be a (Bl2)-point, if in addition the turning point B is tbe improper point of p (see Fig. 46).

, • ,. - - • - ... -- •• - - .. " A'

flrlhogOnalllyperbola

370

371

E. A. DlJKSMAN 54

---Fig. 46

Received Mag 15, 1970

REFERENCES

1. MEYER zen CAPELLES, W .• Fünt- und sechspunktige Gerndtührung in Sonderiagen des ebImen Gelenkvierecks. Forschungsberichte des Wirtscllafts- und Verkehrsminlsterlums Nordrhein-WesUalen, Nr. 481, Westdeutscher Verlag/Köln und Opladen, 1958. 2. VELDKAMP, G. R., Curooture theory in plQne kinema/ies. J. B. Wolters, Groningen, 1963.